For example for we have , note that the coefficients of the multiples of 5 are the maximum powers of 5 dividing these numbers. This comes from - (k,5)=1- So the maximum power of 5 dividing the number is itself multiplied by the maximum such that

Now since it follows that the maximum such that is

Thus:

It may also be written as: ( note that appears in as many sums as the maximum such that )

PS- That's not a factorial

November 15th 2008, 09:11 AM

Soroban

Hello, chandni!

Opalg pointed out a big error in my calculatitons . . . *blush*

Quote:

Find the number of 0's at the end of: .

Every factor of 5 (combined with an even number) produces a final zero.

How many 5's are in that product?

Zeros arise from: .

We are concerned with the factors of 5 (only).

The sum of the series is . . Hence, the product contains a factor of . . . . no

I overlooked these factors: .

. .

My total was short by: .fives.

The product contains a factor of: .

Therefore, the product ends in 1300 zeros.

Thank you, Opalg!
.

November 16th 2008, 07:20 PM

chandni

recheck ur solution

hey i got dat but m confused as to y u included 100^100 and 50^50 in ur calculations the 2nd time wen u had already included it previously...jus chk out dat n do reply